Animated Solution for Physics - Thermodynamics: The root mean square speed of molecules of a given mass of a gas at $27^{\circ }\text{C}$ and $1$ atmosphere pressure is $200{\text{ ms}}^{-1}$. The root mean square speed of molecules of the gas at $127^{\circ }\text{C}$ and $2$ atmosphere pressure is $\frac{x}{\sqrt{3}}{\text{ ms}}^{-1}$. The value of $x$ will be ......... .

Imagine you are observing a sealed container of gas. In the first state, the gas is at a comfortable $27^{\circ }\text{C}$ and $1\text{ atm}$ of pressure. The molecules are zipping around with a root mean square (RMS) speed of $200\text{ m/s}$.

Now, we heat the gas up to $127^{\circ }\text{C}$ and simultaneously increase the pressure to $2\text{ atm}$. The question asks us to find the new RMS speed, which is given in the form $\frac{x}{\sqrt{3}}\text{ m/s}$. Our mission is to find the value of $x$.

Before we dive into calculations, let's look at the master equation for the RMS speed of gas molecules:

Look closely at this equation. Do you see pressure ($p$) anywhere? No! This is a classic trap set by examiners. For an ideal gas, the RMS speed depends only on the absolute temperature ($T$) and the molar mass ($M$). The change in pressure is completely irrelevant to the speed of the molecules as long as we know the temperature.

Since we are dealing with the same gas in both states, the molar mass $M$ is constant. The universal gas constant $R$ is, of course, constant. This means the RMS speed is directly proportional to the square root of the absolute temperature:

To compare the two states, we can set up a simple ratio. But first, we must convert our temperatures from Celsius to Kelvin. This is a crucial step where many students make a silly mistake.

Now, let's set up our ratio:

Substituting the values we know:

Simplifying the fraction inside the square root gives us $\frac{3}{4}$. Taking the square root of that yields:

Cross-multiplying to solve for $v_2$, we get:

The problem states that the new RMS speed is $\frac{x}{\sqrt{3}}\text{ m/s}$. By comparing our result with this expression, it is crystal clear that:

We have successfully navigated the trap and found the correct answer!